I'm not sure I am on the right path. Maybe, I simply need to pick a couple of values to demonstrate that the given relationship is true, but it feels like it might not be enough. In other words, "show" up there looks a bit ambiguous. Any help is highly appreciated.

My friend,why use calculus to get nowhere,when u can use (circular) trigonometry to prove in an elegant way?????????
Take "tan"out both sides of your identity and use what i'll write below:
[tex] \tan(a-b)=:\frac{\sin(a-b)}{\cos(a-b)}=\frac{\sin a\cos b-\sin b\cos a}{\cos a\cos b+\sin a\sin b} [/tex]

Simplify the last fraction through the product of 'cosines' to find the celebrated formula
[tex] \tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b} [/tex]
Substitute in the last identity:
[tex] a\rightarrow \arctan x;b\rightarrow \arctan y [/tex]
,make use of the fact that "tan",on the interval give in the problem is a uniform function and get exactly the formula u would get if taking "tan" from the identity u need to prove.

U didn't. :tongue2: I can't explain in another way the bunch of c*** at numbers (2) and (3)...

Formulas (4) pp.(7) give a proof for the trigonometrical identity that can be used to prove your identity.I stated that proof just to let u know i didn't invent it nor rediscovered it hundreds of years later.

I said to apply tangent on both sides of your identity:
[tex] \tan(\arctan x-\arctan y) =\tan[\arctan(\frac{x-y}{1+xy})] [/tex]

Work the right hand side:
[tex] \tan[\arctan(\frac{x-y}{1+xy})] =\frac{x-y}{1+xy} [/tex] (2)

Interpretation of the relations (1) and (2) is that you have shown that:
[tex] \tan A=C (1') \tan B=C (2') [/tex](3)
From (3) it follows immediately
[tex] \tan A=\tan B [/tex] (4)
From (4) and from tha fact that on the interval [itex] (-\frac{\pi}{2},\frac{\pi}{2}) [/itex] the function "tan" is a uniform/surjective function,u get that
[tex] A=B [/tex] (5)
,which is nothing but your identity in symbolic form.